When is the set of all upper triangular matrices in $\text{GL}(2, \mathbb{Z}/p \mathbb{Z})$ is abelian?

Solution 1:

Let $R$ be a commutative ring, and let $B$ be an abelian group fitting in the chain of inclusions $$\{\pmatrix{a&0\\0&c} : a,c \in R^\times\}=:T \subseteq B \subseteq U:= \{\pmatrix{a&b\\0&c} : a,c \in R^\times, b \in R\}.$$

Conjugation with $\pmatrix{a&0\\0&c}\in T$ maps $\pmatrix{r&b\\0&s}$ to $\pmatrix{r&ac^{-1}b\\0&s}$, so for $B$ to be abelian we need $R^\times \cdot b =b$ for all $b$ such that some $\pmatrix{r&b\\0&s} \in B$.

Hence, for $T \subsetneq B$, all elements $u-1$ for $u \in R^\times$ have to be zero-divisors in $R$. (Indeed, as soon a some element $\pmatrix{1&b\\0&1} \in B$, one similarly shows that $b \cdot (a-c) = 0$ for all units $a,c$, i.e. if $b \neq 0$, each pairwise difference of units must be a zero-divisor in $R$.)

And, as soon as there exists $1 \neq a \in R^\times$, then the full set $U$ of upper triangulars cannot be abelian (take $b=1$ in the above), in fact all elements $b$ appearing in some upper right corner have to be zero-divisors as well.

In particular, let $n \ge 2$. If $R$ contains $\mathbb Z/n$ or $\mathbb Z$, a strict inclusion $T \subsetneq B$ is impossible unless $n$ is even.

If, on the other hand, $n=2m$, and $R=\mathbb Z/n$, then one can check directly that $B=\{\pmatrix{r&b\\0&s}: r,s \in R^\times, b \in m\cdot(\mathbb Z/n) \}$ is indeed abelian. Yours is the case $m=2$.

For $R= \mathbb Z/n$, the full set $U$ is abelian if and only if $n=2$.